使用系统命令:

system("/usr/bin/seq 1000");

这是我的两个解。第一个是c#，第二个是C语言:

C#:

const int limit = 1000;

Action<int>[] actions = new Action<int>[2];
actions[0] = (n) => { Console.WriteLine(n); };
actions[1] = (n) => { Console.WriteLine(n);  actions[Math.Sign(limit - n-1)](n + 1); };

actions[1](0);

C:

#define sign(x) (( x >> 31 ) | ( (unsigned int)( -x ) >> 31 ))

void (*actions[3])(int);

void Action0(int n)
{
    printf("%d", n);
}

void Action1(int n)
{
    int index;
    printf("%d\n", n);
    index = sign(998-n)+1;
    actions[index](++n);
}

void main()
{
    actions[0] = &Action0;
    actions[1] = 0; //Not used
    actions[2] = &Action1;

    actions[2](0);
}

#include <stdio.h>

int show(int i) {
   printf("%d\n",i);
   return( (i>=1000) || show(i+1));
}


int main(int argc,char **argv) {
   return show(1);
}

||操作符使递归调用短路，以显示i为>= 1000时的情况。

#include<iostream>
#include<stdexcept>

using namespace std;

int main(int arg)

{
    try

    {

        printf("%d\n",arg);
        int j=1/(arg-1000);
        main(arg+1);
    }

    catch(...)
    {
        exit(1);
    }
}

您可以使用System()打印1到1000(通过使用DOS命令)

 include <process.h>
 void main()
 {
     system("cmd.exe /c for /l %x in (1, 1, 1000) do echo %x" );
 }

运行程序的.exe(可执行)文件，显示1到1000

注意:在WINDOWS中测试

编译时递归!: P

#include <iostream>
template<int N>
struct NumberGeneration{
  static void out(std::ostream& os)
  {
    NumberGeneration<N-1>::out(os);
    os << N << std::endl;
  }
};
template<>
struct NumberGeneration<1>{
  static void out(std::ostream& os)
  {
    os << 1 << std::endl;
  }
};
int main(){
   NumberGeneration<1000>::out(std::cout);
}